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# Rank and Trace of Idempotent matrix (TIFR 2013 problem 39)

Question:

True/False?

If $$A$$ is a complex nxn matrix with $$A^2=A$$, then rank$$A$$=trace$$A$$.

Hint:

What are the eigenvalues of $$A$$? What is trace in terms of eigenvalues?

Discussion:

If $$v$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$ then $$Av=\lambda v$$, therefore $$\lambda v=Av=A^2v=\lambda Av =\lambda^2 v$$. Therefore, since any eigenvector is non-zero, $$\lambda =0 or 1$$.

Sum of eigenvalues is trace of the matrix. So, trace$$A$$= number of non-zero eigenvalues= total number of eigenvalues – number of 0 eigenvalues

Since $$A$$ satisfies the polynomial $$x^2-x$$, the minimal polynomial is either $$x$$ or $$x-1$$ or $$x(x-1)$$. This means the minimal polynomial breaks into distinct linear factors, so $$A$$ is diagonalizable. Therefore, the algebraic multiplicity of an eigenvalue is same as its geometric multiplicity.

In total there are n eigenvalues (for A is nxn) and the number of 0-eigenvalues is the algebraic multiplicity of 0, which is same as the geometric multiplicity of 0, i.e, the dimension of the kernel of A.

Therefore, trace$$A$$$$=n-$$nullity$$A$$.

By the rank-nullity theorem, the right hand side of the above equation is rank$$A$$.

September 22, 2017